Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving the plane region about the y-axis. y

Answer:

[tex]Volume = \frac{384}{7}\pi[/tex]

Step-by-step explanation:

Given (Missing Information):

[tex]y = x^\frac{3}{2}[/tex]; [tex]y = 8[/tex]; [tex]x=0[/tex]

Required

Determine the volume

Using Shell Method:

[tex]V = 2\pi \int\limits^a_b {p(y)h(y)} \, dy[/tex]

First solve for a and b.

[tex]y = x^\frac{3}{2}[/tex] and [tex]y = 8[/tex]

Substitute 8 for y

[tex]8 = x^\frac{3}{2}[/tex]

Take 2/3 root of both sides

[tex]8^\frac{2}{3} = x^{\frac{3}{2}*\frac{2}{3}}[/tex]

[tex]8^\frac{2}{3} = x[/tex]

[tex]2^{3*\frac{2}{3}} = x[/tex]

[tex]2^2 = x[/tex]

[tex]4 =x[/tex]

[tex]x = 4[/tex]

This implies that:

[tex]a = 4[/tex]

For [tex]x=0[/tex]

This implies that:

[tex]b=0[/tex]

So, we have:

[tex]V = 2\pi \int\limits^a_b {p(y)h(y)} \, dy[/tex]

[tex]V = 2\pi \int\limits^4_0 {p(y)h(y)} \, dy[/tex]

The volume of the solid becomes:

[tex]V = 2\pi \int\limits^4_0 {x(8 - x^{\frac{3}{2}}}) \, dx[/tex]

Open bracket

[tex]V = 2\pi \int\limits^4_0 {8x - x.x^{\frac{3}{2}}} \, dx[/tex]

[tex]V = 2\pi \int\limits^4_0 {8x - x^{\frac{2+3}{2}}} \, dx[/tex]

[tex]V = 2\pi \int\limits^4_0 {8x - x^{\frac{5}{2}}} \, dx[/tex]

Integrate

[tex]V = 2\pi * [{\frac{8x^2}{2} - \frac{x^{1+\frac{5}{2}}}{1+\frac{5}{2}}]\vert^4_0[/tex]

[tex]V = 2\pi * [{4x^2 - \frac{x^{\frac{2+5}{2}}}{\frac{2+5}{2}}]\vert^4_0[/tex]

[tex]V = 2\pi * [{4x^2 - \frac{x^{\frac{7}{2}}}{\frac{7}{2}}]\vert^4_0[/tex]

[tex]V = 2\pi * [{4x^2 - \frac{2}{7}x^{\frac{7}{2}}]\vert^4_0[/tex]

Substitute 4 and 0 for x

[tex]V = 2\pi * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [{4*0^2 - \frac{2}{7}*0^{\frac{7}{2}}])[/tex]

[tex]V = 2\pi * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [0])[/tex]

[tex]V = 2\pi * [{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}][/tex]

[tex]V = 2\pi * [{64 - \frac{2}{7}*2^2^{*\frac{7}{2}}][/tex]

[tex]V = 2\pi * [{64 - \frac{2}{7}*2^7][/tex]

[tex]V = 2\pi * [{64 - \frac{2}{7}*128][/tex]

[tex]V = 2\pi * [{64 - \frac{2*128}{7}][/tex]

[tex]V = 2\pi * [{64 - \frac{256}{7}][/tex]

Take LCM

[tex]V = 2\pi * [\frac{64*7-256}{7}][/tex]

[tex]V = 2\pi * [\frac{448-256}{7}][/tex]

[tex]V = 2\pi * [\frac{192}{7}][/tex]

[tex]V = [\frac{2\pi * 192}{7}][/tex]

[tex]V = \frac{\pi * 384}{7}[/tex]

[tex]V = \frac{384}{7}\pi[/tex]

Hence, the required volume is:

[tex]Volume = \frac{384}{7}\pi[/tex]

The volume of the solid of revolution generated by revolving the plane region about the y-axis is [tex]\frac{10240\pi}{3}[/tex] cubic units.

Let be [tex]f(x) = x^{5/2}[/tex] and [tex]g(x) = 32[/tex], whose point of intersection is [tex](x,y) = (4, 32)[/tex]. Solid of revolution generated by revolving the plane region about the y-axis is defined by the following formula:

[tex]V = 2\pi\int\limits^{32}_0 {y\cdot g(y)} \, dy[/tex] (1)

If we know that [tex]g(y) = y^{2/5}[/tex], then the volume of the solid of revolution is:

[tex]V = 2\pi\int\limits^{32}_{0} {y^{7/5}} \, dy[/tex]

[tex]V = 2\pi\cdot \left(\frac{5}{12}\cdot y^{12/5}\right)\left|_{0}^{32}[/tex]

[tex]V = \frac{5\pi}{6} \cdot y^{12/5}|\limit_{0}^{32}[/tex]

[tex]V = \frac{5\pi}{6}\cdot 32^{12/5}[/tex]

[tex]V = \frac{10240\pi}{3}[/tex]

The volume of the solid of revolution generated by revolving the plane region about the y-axis is [tex]\frac{10240\pi}{3}[/tex] cubic units. [tex]\blacksquare[/tex]

The statement is incomplete. Complete form is shown below:

Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving the plane region about the y-axis. [tex]y = x^{5/2}[/tex], [tex]y = 32[/tex], [tex]x = 0[/tex].

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